Air enters a 16-cm-diameter pipe steadily at 200 kPa and 20°C with a velocity of 5 m/s. Air is heated as it flows, and it leaves the pipe at 180 kPa and 43°C. The gas constant of air is 0.287 kPa·m3/kg·K. Whats the volumetric flow rate of the inlet/outlet, mass flow rate and velocity & volume flow rate at the exit?

Answer:

correct answer is (b)  Longitudinal lines remain straight

Explanation:

solution

As we know that Deformation of circular shaft in the torsion is associate with twisting of  shaft more than an specify with the yielding limit.

so when any angle of twist is obtain in the torsion and that is beyond the specified safety limit of shaft

than that shaft will be fail.

but it does not regain its original shape and it will cause permanent deformation

so that we can say longitudinal lines which is twist, they will not regain to original back position as straight

but they will remain in curved shape.

so here incorrect statement is b Longitudinal lines remain straight

Though torsion theory assumes that cross sections of a torsionally loaded shaft remain flat, in reality, under heavy loading, they can warp and so this statement is not entirely accurate.

When a circular shaft deforms under torsion, certain assumptions are made about its deformation according to the torsion theory. These assumptions include:

1) cross sections remain flat and perpendicular to the axis of the shaft,

2) longitudinal lines remain straight,

3) circular sections remain circular, and

4) radial lines on the sections remain straight. However, the statement that is not entirely accurate is that cross sections remain flat. In reality, under severe torsional loading, the cross sections might warp and not remain entirely flat.

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Answer:

The centripetal acceleration at highway speed is greater.

Explanation:

We assume the motion of the car is uniformly accelerated. Let the highway speed be v.

By the equation of motion,

[tex]v=u+at[/tex]

[tex]a=\dfrac{v-u}{t}[/tex]

u is the initial velocity, a is acceleration and t is time

Because the car starts from rest, u = 0.

[tex]a_T=\dfrac{v}{t}[/tex]

This is the tangential acceleration of the thread of the tire.

The centripetal acceleration is given by

[tex]a_C=\dfrac{v^2}{r}[/tex]

r is the radius of the tire.

Comparing both accelerations and applying commonly expected values to r and t, the centripetal acceleration is seen to be greater. The radius of a tyre is, on the average, less than 0.4 m. Then the centripetal acceleration is about

[tex]a_C=\dfrac{v^2}{0.3}=2.5v^2[/tex]

The tangential acceleration can only be greater in the near impossible condition that the time to attain the speed is on the order of microseconds.

Answer:

The sum of the kinetic energies of the alpha particle and the new nucleus = (1.359098 × 10⁻¹²) J

Explanation:

We will use the conservation of energy theorem for extremely small particles,

Total energy before split = total energy after split

That is,

Total energy of the original nucleus = (total energy of the new nucleus) + (total energy of the alpha particle)

Total energy of these subatomic particles is given as equal to (rest energy) + (kinetic energy)

Rest energy = mc² (Einstein)

Let Kinetic energy be k

Kinetic energy of original nucleus = k₀ = 0 J

Kinetic energy of new nucleus = kₙ

Kinetic energy of alpha particle = kₐ

Mass of original nucleus = m₀ = (3.969554 × 10⁻²⁵) kg

Mass of new nucleus = mₙ = (3.902996 × 10⁻²⁵) kg

Mass of alpha particle = mₐ = (6.640678 × 10⁻²⁷) kg

Speed of light = (2.9979246 × 10⁸) m/s

Total energy of the original nucleus = m₀c² (kinetic energy = 0, since it was originally at rest)

Total energy of new nucleus = (mₙc²) + kₙ

Total energy of the alpha particle = (mₐc²) + kₐ

(m₀c²) = (mₙc²) + kₙ + (mₐc²) + kₐ

kₙ + kₐ = (m₀c²) - [(mₙc²) + (mₐc²)

(kₙ + kₐ) = c² (m₀ - mₙ - mₐ)

(kₙ + kₐ) = (2.9979246 × 10⁸)² [(3.969554 × 10⁻²⁵) - (3.902996 × 10⁻²⁵) - (6.640678 × 10⁻²⁷)]

(kₙ + kₐ) = (8.98755191 × 10¹⁶)(1.5122 × 10⁻²⁹) = (1.35909760 × 10⁻¹²) J

Answer:

Normal Conversation: i=106i0

i(dB)=60

Power saw a 3 feet: i=1011i0

i(dB)=110

Jet engine at 100 feet: i=1018i0

i(dB)=180

Explanation:

if these are the same as edge, then these are the answers! :)

Normal Conversation: i=106i0

The intensity in decibels is 60,110,180, respectively.

⇒ normal conversation: i =60(dB)

⇒power saw 3 feet: i =110(dB)

⇒jet engine at 100 feet: i = 180(dB)

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Answer:

    a = 4.8 10⁻⁵ m

Explanation:

The diffraction phenomenon is described by the expression

        a sin θ = m λ

How the pattern is observed on a distant screen

        tan θ = y / L = sin θ / cos θ

Since the angle is very small in these experiments ’we can approximate the tangent function

    tan θ = sin θ = y / L

We substitute

           a y / L = m λ

The first minimum occurs for m = 1

        a = λ L / y

        a = 680 10⁻⁹ 5.5 / 0.078

        a = 4.8 10⁻⁵ m

Yes, the cart will move to the left after the ball bounces back from the surface. Thus, option (B) is correct.

When you throw the ball at the vertical surface attached to the cart, the ball's momentum changes due to the collision. Initially, both you and the cart are at rest, so the total momentum of the system (you + cart + ball) is zero.

As the ball collides with the vertical surface and bounces back, it changes its direction and gains momentum in the opposite direction. This change in momentum is due to the impulse imparted to the ball during the collision.

According to the law of conservation of momentum, when the ball bounces back from the vertical surface, it exerts a backward force on the cart due to the change in momentum.

Since momentum is conserved, the cart will experience an equal and opposite forward force, causing it to move to the left.

Thus, option (B) is correct.

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The cart will move due to conservation of momentum. When the ball bounces back the cart moves to the left following Newton's Third Law of Motion.

This question relates to the law of conservation of momentum in the realm of physics. When you throw a ball towards a vertical surface on the cart, both the ball and the cart will experience a change in momentum. As the ball bounces back (with momentum in the opposite direction the law of conservation of momentum applies the total momentum before the event (throwing the ball) must equal the total momentum after the event (ball bouncing back).

The answer to the scenario is B. Yes and it moves to the left. When the ball hits the surface and bounces back it applies a force to the cart and due to Newton's third law the cart will move in the opposite direction hence to the left assuming the ball was thrown to the right.

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The tip of one prong of a tuning fork undergoes SHM of frequency 1000 Hz and amplitude 0.40 mm. For this tip, what is the magnitude of the

(a) maximum acceleration,

(b) maximum velocity,

(c) acceleration at tip displacement 0.20 mm, and

(d) velocity at tip displacement 0.20 mm?

(a) 15795.5m/s²

(b) 2.5m/s

(c) 7897.7 m/s²

(d) 2.2m/s

The displacement, y, of a body undergoing simple harmonic motion (SHM) is given by

y = A sin (ωt + φ)             ------------------(i)

Where;

A = maximum displacement or amplitude of the body

ω = angular frequency of the body

t = time taken for the displacement

φ = phase constant

The velocity, v, of the body can be found by differentiating equation (i) as follows;

v = Aω cos (ωt + φ)          ------------------(ii)

Where;

Aω = maximum velocity or amplitude of the velocity of the body

Also, the acceleration of the body can be found by differentiating equation (ii) as follows;

a = -Aω² sin(ωt + φ)                  --------------------(iii)

Where;

-Aω² = maximum acceleration or amplitude of the acceleration of the body

(a) From equation (iii), the magnitude of the maximum acceleration [tex]a_{max}[/tex] is given by;

[tex]a_{max}[/tex] = Aω²           ----------------(iv)

Where;

A = amplitude = 0.40mm = 0.00040m

ω = 2 π f            [Take π = 3.142. Also, f = frequency of the motion = 1000Hz]

=> ω = 2 x 3.142 x 1000 = 6284 rad/s

Substitute these values into equation (iv) as follows;

[tex]a_{max}[/tex] = 0.00040 x 6284² = 15795.5m/s²

Therefore, the magnitude of the maximum acceleration is 15795.5m/s²

========================================================

(b) From equation (ii), the magnitude of the maximum velocity [tex]v_{max}[/tex], is given by;

[tex]v_{max}[/tex] = Aω          ----------------(v)

Where;

A = amplitude = 0.40mm = 0.00040m

ω = 2 π f            [Take π = 3.142. Also, f = frequency of the motion = 1000Hz]

=> ω = 2 x 3.142 x 1000 = 6284 rad/s

Substitute these values into equation (v) as follows;

[tex]v_{max}[/tex] = 0.00040 x 6284 = 15795.5m/s²

Therefore, the magnitude of the maximum velocity is 2.5m/s

========================================================

(c) Comparing equations (i) and (iii),  equation (iii) can be written as;

a = -ω² y            -------------------(vi)

Therefore, to get the acceleration at tip displacement of 0.20mm, substitute y = 0.20mm = 0.00020m and ω = 6284rad/s into equation (vi) as follows;

a = - 6284² x 0.00020

a = - 7897.7 m/s²

Therefore, the magnitude of the acceleration at the tip displacement is 7897.7 m/s²

========================================================

(d) Recall that;

sin²θ + cos²θ = 1

=> cos²θ = 1 - sin²θ

=> cosθ = √(1 - sin²θ )

=> cos (ωt + φ)  = √(1 - sin² (ωt + φ))

Substitute this value into equation (ii) as follows;

v = Aω √(1 - sin² (ωt + φ))

v = ω√(A² - A²sin² (ωt + φ))              

Now, comparing the equation above and equation (i), the equation above can be written as;

v = ω√(A² - y²)           -------------(vii)

Therefore, to get the velocity at tip displacement of 0.20mm, substitute y = 0.20mm = 0.00020m, ω = 6284rad/s and A = 0.00040m into equation (vii) as follows;

v = 6284√(0.00040² - 0.00020²)

v = 6284√(0.00000060)

v = 2.2m/s

Therefore, the magnitude of the velocity at the tip displacement is 2.2 m/s

Answer:

i = f = 0.1 m until the lens moves towards the screen 0.1 m

Explanation:

For this exercise let's use the constructor equation

         1 / f = 1 / o + 1 / i

Where f is the focal distance, or the distance to the object and "i" the distance to the image

It indicates that the focal distance is 100 mm (f = 100 mm), when an object is at infinity the image is formed at its focal length

     1 / f = 1 / inf + 1 / i = 1 / i

      i = f = 100 mm

At this point the screen is placed

For the shortest distance the lens has to move a little so the distance to the image is

        f = 100 mm = 0.1 m

        o = 6 pm = 6 10⁻¹² m

        i = 100 + x

        1 / f = 1 / o + 1 / 100+ x

        1 /( 0.10 + x) = 1 / f - 1 / o

        1 / 0.100+ x = 1/0.100 - 1/6 10-12

        1 / 0.100 + x = 10 - 10¹¹ = -10¹¹

        0.100 + x = -10⁻¹¹

        x = -10⁻¹¹ -10⁻¹

        x = -10⁻¹

        x = - 0.1 m

This negative distance indicates that the lens moves towards the screen 0.1 m

Answer:

The average current that this cell phone draws when turned on is 0.451 A.

Explanation:

Given;

voltage of the phone, V = 3.7 V

electrical energy of the phone battery, E = 3.15 x 10⁴ J

duration of battery energy, t = 5.25 h

The power the cell phone draws when turned on, is the rate of energy consumption, and this is calculated as follows;

[tex]P = \frac{E}{t}[/tex]

where;

P is power in watts

E is energy in Joules

t is time in seconds

[tex]P = \frac{3.15*10^4}{5.25*3600s} = 1.667 \ W[/tex]

The average current that this cell phone draws when turned on:

P = IV

[tex]I = \frac{P}{V} =\frac{1.667}{3.7} = 0.451 \ A[/tex]

Therefore, the average current that this cell phone draws when turned on is 0.451 A.

The average current that the cell phone draws when turned on is approximately [tex]\(0.450 \, \text{A}\)[/tex].

To find the average current drawn by the cell phone, we need to use the relationship between power, voltage, and current. Let's go step by step.

Step 1: Calculate the power consumed by the cell phone

We already have:

- The total energy provided by the battery: [tex]\( \Delta W = 3.15 \times 10^4 \, \text{J} \)[/tex]

- The total time of operation: [tex]\( \Delta t = 5.25 \, \text{hours} \)[/tex]

First, convert the operation time from hours to seconds:

[tex]\[\Delta t = 5.25 \, \text{hours} \times 3600 \, \text{seconds/hour} = 5.25 \times 3600 = 18900 \, \text{s}\][/tex]

Now, calculate the power P using the energy and time:

[tex]\[P = \frac{\Delta W}{\Delta t} = \frac{3.15 \times 10^4 \, \text{J}}{18900 \, \text{s}} \approx 1.666 \, \text{W}\][/tex]

Step 2: Use the power to find the average current

We know that power P, voltage V, and current I are related by the equation:

[tex]\[P = VI\][/tex]

We can solve for the current I

[tex]\[I = \frac{P}{V}\][/tex]

Given the battery voltage [tex]\(V = 3.70 \, \text{V}\)[/tex] and the power [tex]\(P \approx 1.666 \, \text{W}\)[/tex], we can calculate the current:

[tex]\[I = \frac{1.666 \, \text{W}}{3.70 \, \text{V}} \approx 0.450 \, \text{A}\][/tex]

The complete question is this:

The battery for a certain cell phone is rated at 3.70 V. According to the manufacturer it can produce 3.15 x 104 J of electrical energy, enough for 5.25 h of operation, before needing to be recharged. Find the average current that this cell phone draws when turned on.

Ok, so what I did so far was convert time into seconds and found Power:

t = 18900 s

P = ΔW/Δt =student submitted image, transcription available below= 1.6666 W

I think you have to use the problem : P = VabI = I2R = εI - I2R

Answer:

λ= 5.24 × 10 ⁻² nC/cm

Explanation:

Given:

distance r = 4.10 cm = 0.041 m

Electric field intensity E = 2300 N/C

K = 9 x 10 ⁹ Nm²/C

To find λ = linear charge density = ?

Sol:

we know that E= 2Kλ / r

⇒ λ = -E r/2K         (-ve sign show the direction toward the wire)

λ = (- 2300 N/C × 0.041 m) / 2 ×  9 x 10 ⁹ Nm²/C

λ = 5.24 × 10 ⁻⁹ C/m

λ = 5.24 nC/m = 5.24 nC/100 cm

λ= 5.24 × 10 ⁻² nC/cm

Answer:

Magnetic field at point having a distance of 2 cm from wire is 6.99 x 10⁻⁶ T

Explanation:

Magnetic field due to finite straight wire at a point perpendicular to the wire is given by the relation :

[tex]B=\frac{\mu_{0}I }{2\pi R }\times\frac{L}{\sqrt{L^{2}+R^{2} } }[/tex]      ......(1)

Here I is current in the wire, L is the length of the wire, R is the distance of the point from the wire and μ₀ is vacuum permeability constant.

In this problem,

Current, I = 0.7 A

Length of wire, L = 0.62 m

Distance of point from wire, R = 2 cm = 2 x 10⁻² m = 0.02 m

Vacuum permeability, μ₀ = 4π x 10⁻⁷ H/m

Substitute these values in equation (1).

[tex]B=\frac{4\pi\times10^{-7}\times 0.7 }{2\pi \times0.02 }\times\frac{0.62}{\sqrt{(0.62)^{2}+(0.02) ^{2} } }[/tex]

B = 6.99 x 10⁻⁶ T

Answer:

0.208 N

Explanation:

Parameters given:

Current, I = 2A

Angle, A = 60°

Magnetic field strength, B = 0.2 T

Length, L = 0.6 m

Magnetic force is given as:

F = I * L * B * sinA

F = 2 * 0.6 * 0.2 * sin60

F = 0.24 * sin60.

F = 0.208 N

Answer:

0.20784 N

Explanation:

The force on conductor carrying current in a magnetic Field is given as,

F = BILsinθ..................... Equation 1

Where F = force on wire carrying current, B = magnetic Field, I = current, L = Length of the wire, θ = angle between the wire and the magnetic Field.

Given: B = 0.2 T, I = 2 A, L = 0.6 m, θ = 60°

Substitute into equation 1

F = 0.2(2)(0.6)sin60°

F = 0.24×0.8660

F = 0.20784 N

Answer:

The answers to the questions are;

(a) The velocity of the truck right after the collision is 20.884 m/s

(b) The change in mechanical energy of the car truck system in the collision is -9076.4384 J

(c) The change in mechanical energy is due to energy consumed by the collision process.

Explanation:

(a) From the principle of conservation of linear momentum, we have

m₁·v₁+m₂·v₂ = m₁·v₃ + m₂·v₄

Where:

m₁ = Mass of the car = 1225.0 kg

m₂ = Mass of the truck = 9700.0 kg

v₁ = Initial velocity of the car = 25.000 m/s

v₂ = Initial velocity of the truck = 20.000 m/s

v₃ = Final velocity of the car right after collision = 18.000 m/s

v₄ = Final velocity of the truck right after collision

Therefore

1225.0 kg × 25.000 m/s  +  9700.0 kg × 20.000 m/s = 1225.0 kg × 18.000 m/s  + 9700.0 kg × v₄

That is 30625 kg·m/s + 194000 kg·m/s = 22050 kg·m/s + 9700.0 kg × v₄

Making v₄ the subject of the formula yields

v₄ = (202575 kg·m/s)÷9700.0 kg = 20.884 m/s

The velocity of the truck right after the collision to five significant figures = 20.884 m/s

(b) The change in mechanical energy of the car truck system in the collision can be found by

The change in kinetic energy of the car truck system

Change in kinetic energy, ΔK.E. = Sum of final kinetic energy - Sum of initial kinetic energy

That is ΔK.E. = ∑ Final K.E -∑ Initial K.E.

ΔK.E. = [tex](\frac{1}{2} m_1v_3^{2}+\frac{1}{2} m_2v_4^{2}) - (\frac{1}{2} m_1v_1^{2} +\frac{1}{2} m_2v_2^{2} )[/tex]

         = ([tex]\frac{1}{2}[/tex]·1225·18²+ [tex]\frac{1}{2}[/tex]·9700·20.884²) - ([tex]\frac{1}{2}[/tex]·1225·25²+[tex]\frac{1}{2}[/tex]·9700·20²)

         = 2313736.0616 kg·m²/s² - 2322812.5 kg·m²/s² =  -9076.4384 kg·m²/s²

1 kg·m²/s² = 1 J ∴ -9076.4384 kg·m²/s² = -9076.4384 J

(c) The energy given off by way of the 9076.4384 J is energy transformed into other forms including

1) Frictional resistance between the tires and the road for the truck and car

2) Frictional resistance in the transmission system of the truck to increase its velocity

3) Sound energy, loud sound heard during the collision

4) Energy absorbed when the car and the truck outer frames are crushed

5) Heat energy in the form of raised temperatures at the collision points of the car and the truck.

6) Energy required to change the velocity of the car over a short distance.

(a) The velocity of the truck right after the collision is 20.95258 m/s east. (b)The change in mechanical energy of the car-truck system in the collision is -2729.87 J. (c) This reduction in mechanical energy is due to energy conversion into heat, sound, and deformation.

Let's analyze the collision using the principles of conservation of momentum and mechanical energy.

The law of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision.

Before the Collision:

After the Collision:

Total momentum after the collision: 22050.0 kg·m/s + (9700.0 kg)(vt)

Setting initial and final momenta equal:

224625.0 kg·m/s = 22050.0 kg·m/s + 9700.0 kg * vt

Solving for vt:

vt = (224625.0 kg·m/s - 22050.0 kg·m/s) / 9700.0 kg = 20.952577 m/s

Therefore, the velocity of the truck right after the collision is 20.95258 m/s (east).

Kinetic energy is calculated using the formula KE = (1/2)mv².

Before the Collision:

After the Collision:

Change in mechanical energy: 2320082.5 J - 2322812.5 J = -2729.87 J

The mechanical energy of the system decreases by 2729.87 J.

The decrease in mechanical energy is due to some energy being converted into other forms such as heat, sound, and deformation of the vehicles during the collision.

Answer:

[tex]F_{cp}=3947.84N[/tex]

Explanation:

From the formula for centripetal force and acceleration we can deduce that:

[tex]F_{cp}=ma_{cp}=m\frac{v^2}{r}=m\frac{(r\omega)^2}{r}=mr\omega^2=mr\omega^2[/tex]

Since one revolution is [tex]2\pi\ rad[/tex], 0.5 revolutions are [tex]\pi\ rad[/tex], so we have:

[tex]F_{cp}=(4kg)(100m)(\pi\ rad/s)^2=3947.84N[/tex]

Answer:

[tex]44,000 Nm^2/C[/tex]

Explanation:

The electric flux through a certain surface is given by (for a uniform field):

[tex]\Phi = EA cos \theta[/tex]

where:

E is the magnitude of the electric field

A is the area of the surface

[tex]\theta[/tex] is the angle between the direction of the field and of the normal to the surface

In this problem, we have:

[tex]E=1.1\cdot 10^4 N/C[/tex] is the electric field

L = 2.0 m is the side of the sheet, so the area is

[tex]A=L^2=(2.0)^2=4.0 m^2[/tex]

[tex]\theta=0^{\circ}[/tex], since the electric field is perpendicular to the surface

Therefore, the electric flux is

[tex]\Phi =(1.1\cdot 10^4)(4.0)(cos 0^{\circ})=44,000 Nm^2/C[/tex]

The electric flux through the sheet will be "44,000 Nm²/C".

An electric field seems to be an area of space that surrounds an electrically charged particle as well as object whereby an electric charge will indeed experience attraction.

According to the question,

Magnitude of electric field, E = 1.1 × 10⁴ N/C

Length, L = 2.0 m

We know,

The area will be:

→ A = L²

By substituting the value,

      = (2.0)²

      = 4.0 m²

hence,

The electric flux will be:

→ [tex]\Phi[/tex] = EA Cosθ

By substituting the values,

      = [tex](1.1.10^4)(4.0)(Cos 0^{\circ})[/tex]

      = 44,000 Nm²/C

Thus the above answer is appropriate.  

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Explanation:

The given data is as follows.

           mass (m) = 5 kg

       Height of tower = 15 m

        u = 7 m/s

    air resistance = 610 v

(a)   Now, differential equation for the given mass which is thrown vertically upwards is as follows.

              [tex]m \frac{d^{2}x}{dt^{2}}[/tex] = F

                  -bv = Fr

Here, mg is downwards due to the force of gravity.

              [tex]\frac{md^{2}x}{dt^{2}} = bv - mg[/tex]

       [tex]\frac{md^{2}x}{dt^{2}} + b \frac{dx}{dt} + mg[/tex] = 0

Hence, the differential equation required to solve the problem is as follows.

       [tex]\frac{md^{2}x}{dt^{2}} + b \frac{dx}{dt} + mg[/tex] = 0    

(b)   When final velocity of the object is equal to zero then the object will reach towards its maximum height and it will start to fall downwards.

              F = [tex]\frac{md^{2}x}{dt^{2}}[/tex]

                 = 0

Therefore, the object reach its maximum height at v = 0.

c. The object with the higher initial temperature will have a greater temperature change when brought into thermal contact with another object.

The object with the higher initial temperature will have a greater temperature change when brought into thermal contact with another object. This is because energy transfers from the object with the higher temperature to the object with the lower temperature until they reach thermal equilibrium.

Mass and specific heat do not directly affect the temperature change. The mass only affects the amount of energy transferred, while the specific heat determines how much energy is needed to raise the temperature.

Therefore, the correct answer is c. The one with the higher initial temperature.

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Answer:

Check attachment for complete questions, the question is not complete

Explanation:

Check attachment for solution

Complete Question

The complete Question is shown on the first and second uploaded image

Answer:

The speed at which they need to push the mass is v = 13.1 m/s

Explanation:

In order to solve this problem we need to consider conservation of energy when the block is at the top of the inclined plane and also when it is on top of the loop

Now Applying the law of conservation of energy

        [tex]mg (2R) + \frac{1}{2} mv^2 = \frac{1}{2} mv_{top}^2 + mg(2R)[/tex]

  where   [tex]mg (2R)[/tex] is potential energy and [tex]\frac{1}{2} mv^2[/tex] is kinetic energy

  and [tex]v_{top}[/tex] is the velocity at the top inclined plane and the top of the loop

        Now considering the formula

                           [tex]\frac{1}{2} mv^2 = \frac{1}{2} mv_{top}^2[/tex]

                            [tex]v^2 = v_{top}^2[/tex]

                            [tex]v = v_{top}[/tex]

Now to obtain [tex]v_{top}[/tex]

   Looking at the question we can say that the centripetal force that made the block move around loop without leaving the track is q=equivalent to the centripetal force  so we have

            [tex]mg = \frac{mv_{top}^2}{R}[/tex]

The m would cancel out each other then cross- multiplying

             [tex]gR = v^2_{top}[/tex]

         [tex]v_{top} = \sqrt{gR}[/tex]

                 [tex]= \sqrt{(9.8 m/s^2)(17.4\ m)}[/tex]

                [tex]= 13.05 m/s[/tex]

                [tex]\approx 13.1 m/s[/tex]

Answer:

The diagram is in the attachment

Explanation:

An ammeter is use to know the current flowing in a circuit,

A voltmeter is use to know the potential difference across an element.

The ideal voltmeter and the ideal ammeter has zero internal resistance, so as to drop as little voltage as possible as current flows through it.

1. Let analyse the first circuit i.e the ammeter connection

The ammeter is connected rightly and all the current coming from the battery will flow into the bulb and the bulb will glow bright using only the current from the battery and the ammeter work in the circuit is only to measure the current from the battery.

Now let analyse the second circuit, the voltmeter connection.

This is a wrong connection and if this is done it will act has high resistance to the current flow. The connecting of voltmeter in series is equivalent to connecting a very high resistance in series with the circuit. By this only small insignificant amount of current flow through the circuit and nearly results in an open circuit.

Conclusion,

The first connection of ammeter bulb will glow brightly because it uses up all the current from the battery but for the voltmeter connection the current has been reduced due to the high resistance of voltmeter and thus reduces current.

Answer:

The maximum energy that can be stored in the capacitor is  6.62 x 10⁻⁵ J

Explanation:

Given;

dielectric constant k = 5.5

the area of each plate, A = 0.034 m²

separating distance, d =  2.0 mm = 2 x 10⁻³ m

magnitude of the electric field =  200 kN/C

Capacitance of the capacitor is calculated as follows;

[tex]C = \frac{k \epsilon A}{d} = \frac{5.5*8.85*10^{-12}*0.034}{2*10^{-3}} = 8.275 *10^{-10} \ F[/tex]

Maximum potential difference:

V = E x d

V = 200000 x 2 x 10⁻³ = 400 V

Maximum energy that can be stored in the capacitor:

E = ¹/₂CV²

E = ¹/₂ x 8.275 x 10⁻¹⁰ x (400)²

E = 6.62 x 10⁻⁵ J

Therefore, the maximum energy that can be stored in the capacitor is  6.62 x 10⁻⁵ J

Answer:

[tex]5.3\times 10^3 N/C[/tex]

Explanation:

We are given that

Distance between plates=d=2.2 cm=[tex]2.2\times 10^{-2} m[/tex]

[tex]1 cm=10^{-2} m[/tex]

[tex]\sigma=47nC/m^2=47\times 10^{-9}C/m^2[/tex]

Using [tex]1 nC=10^{-9} C[/tex]

We have to find the magnitude of E in the region between the plates.

We know that the electric field for parallel plates

[tex]E=\frac{\sigma}{2\epsilon_0}[/tex]

[tex]E_1=\frac{\sigma}{2\epsilon_0}[/tex]

[tex]E_2=\frac{\sigma}{2\epsilon_0}[/tex]

[tex]E=E_1+E_2[/tex]

[tex]E=\frac{\sigma}{2\epsilon_0}+\frac{\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}[/tex]

Where [tex]\epsilon_0=8.85\times 10^{-12}C^2/Nm^2[/tex]

Substitute the values

[tex]E=\frac{47\times 10^{-9}}{8.85\times 10^{-12}}[/tex]

[tex]E=5.3\times 10^3 N/C[/tex]

Hence, the magnitude of E in the region between the plates=[tex]5.3\times 10^3 N/C[/tex]

The magnitude of the electric field E in the region between two parallel conducting plates carrying opposite charges of equal magnitude can be calculated using the formula E = σ / ε0. Given the surface charge density σ of 47.0 nC/m2 and the permittivity of free space ε0 as 8.85 × 10-12 F/m, the electric field E results as 5.31 × 103 N/C.

The system mentioned in the question is known as a parallel-plate capacitor. It is important to understand that the electric field between two large parallel conducting plates carrying opposite charges of equal magnitude is perpendicular to the plates and is constant in both magnitude and direction. The electric field E between the plates of a capacitor can be calculated by the formula E = σ / ε0, where σ is the surface charge density on one plate, and ε0 is the permittivity of free space, which is approximately 8.85 × 10-12 F/m.

Given in the question, the surface charge density σ is 47.0 nC/m2. To convert nanocoulombs to coulombs, multiply by 10-9, so σ = 47.0 × 10-9 C/m2. Applying the values to the formula, we get E = (47.0 × 10-9 C/m2) / (8.85 × 10-12 F/m) = 5.31 × 103 N/C. Therefore, the electric field E in the region between the plates is 5.31 × 103 N/C.

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Answer:

a) b) d)

Explanation:

The question is incomplete. The Complete question might be

In an inertial frame of reference, a series of experiments is conducted. In each experiment, two or three forces are applied to an object. The magnitudes of these forces are given. No other forces are acting on the object. In which cases may the object possibly remain at rest? The forces applied are as follows: Check all that apply.

a)2 N; 2 N

b) 200 N; 200 N

c) 200 N; 201 N

d) 2 N; 2 N; 4 N

e) 2 N; 2 N; 2 N

f) 2 N; 2 N; 3 N

g) 2 N; 2 N; 5 N

h ) 200 N; 200 N; 5 N

For th object to remain at rest, sum of all forces must be equal to zero. Use minus sign to show opposing forces

a) 2+(-2)=0 here minus sign is to show the opposing firection of force

b) 200+(-200)=0

c) 200+(-201)[tex]\neq[/tex]0

d) 2+2+(-4)=0

e) 2+2+(-2)[tex]\neq[/tex]0

f) 2+2+(-3) [tex]\neq[/tex]0; 2+(-2)+3[tex]\neq[/tex]0

g) 2+2+(-5)[tex]\neq[/tex]0; 2+(-2)+5[tex]\neq[/tex]0

h)200 + 200 +(-5)[tex]\neq[/tex]0; 200+(-200)+5[tex]\neq[/tex]0

Explanation:

According to Newton's second law of motion,

            [tex]m_{1}g - T = m_{1}a[/tex] ......... (1)

and,    [tex]T - m_{2}g = m_{2}a[/tex] ......... (2)

When we add both equations, (1) and (2) then the expression obtained for "a" is as follows.

             a = [tex]\frac{m_{1} - m_{2}}{m_{1} + m_{2}} \times g[/tex]

                = [tex]\frac{17.7 - 11.1}{17.7 + 11.1} \times 9.8[/tex]

                = [tex]\frac{6.6}{28.8} \times 9.8[/tex]

                = 2.24 [tex]m/s^{2}[/tex]

Now, putting the value of "a" in equation (1) then we will calculate the tension as follows.

              [tex]m_{1}g - T = m_{1}a[/tex]

              [tex]17.7 \times 9.8 - T = 17.7 \times 2.24[/tex]

                   173.46 - T = 39.648

                       T = 133.812 N

Thus, we can conclude that the magnitude of their acceleration is 2.24 [tex]m/s^{2}[/tex] and the tension T is 133.812 N in the rope.

Answer:

Explanation:

m1 = 17.7 kg

m2 = 11.1 kg

Let a be the acceleration and T be the tension in the string.

use Newton's second law

m1 g - T = m1 x a ....(1)

T - m2 g = m2 x a ..... (2)

Adding both the equations

(m1 - m2) g = ( m1 + m2 ) x a

(17.7 - 11.1 ) x 9.8 = (17.7 + 11.1) x a

64.68 = 28.8 a

a = 2.25 m/s²

Put the value of a in equation (1)

17.7 x 9.8 - T = 17.7 x 2.25

173.46 - T = 39.825

T = 133.64 N

Answer:

3,4,1,2

Explanation:

The filing procedure involves Filing in terms of the first letter.If the first letters are common, file in terms of the second letter then File in terms of surnames.If the surnames are common, file in terms of the initial. If the surname is more than one then file under the first surname etc.

Answer:

6104 N/C.

Explanation:

Given:

k = 8.99 × 10^9 Nm2/C^2

Qx = 1.3 × 10^-5 C

rx = 7 m

Qy = 1 × 10−5 C

ry = 4 m

E = F/Q

= kQ/r^2

Ex = (8.99 × 10^9 × 1.3 × 10^−5) ÷ 7^2

= 2385.1 N/C.

Ey = (8.99 × 10^9 × 1.0 × 10^−5) ÷ 4^2

= 5618.75 N/C

Eo = sqrt(Ex^2 + Ey^2)

= sqrt(3.157 × 10^7 + 5.69 × 10^6)

= 6104 N/C.

The magnitude of the electric field at the origin due to the charges fixed at (7 m, 0) and (0, 4 m) is approximately 6107.83 N/C. This was calculated by determining the electric fields from each charge and then using vector addition to find the resultant electric field. The calculation involved using the formula E = k |q| / r² for each charge, followed by determining the resultant using Pythagoras' theorem.

Let's determine the electric field at the origin due to two fixed charges. The first charge is -1.3 × 10⁻⁵ C located at (7 m, 0), and the second charge is 1 × 10⁻⁵C located at (0, 4 m).

The electric field due to a point charge q at a distance r is given by E = k |q| / r², where k = 8.99 × 10⁹ N·m²/C² is the Coulomb constant.

Step-by-Step Calculation

Calculate the distance from each charge to the origin:

Charge 1 at (7 m, 0): r₁ = 7 m

Charge 2 at (0, 4 m): r₂ = 4 m

Compute the electric field due to each charge:

For q₁ = -1.3 × 10⁻⁵ C:
E₁ = k |q₁| / r₁² = (8.99 × 10⁹ N·m²/C²) (1.3 × 10⁻⁵ C) / (7 m)² ≈ 2381.57 N/C along the negative x-direction.

For q₂ = 1 × 10⁻⁵ C:
E₂ = k |q₂| / r₂² = (8.99 × 10⁹ N·m²/C²) (1 × 10⁻⁵ C) / (4 m)² ≈ 5621.88 N/C along the positive y-direction.

Determine the resultant electric field at the origin using vector addition. Since the fields are perpendicular:

Resultant E = √(E₁² + E₂²)
≈ √((2381.57 N/C)² + (5621.88 N/C)²)
≈ √(5662179.6649 + 31643689.7344)
≈ √(37305869.3993)
Resultant E ≈ 6107.83 N/C

The magnitude of the resultant electric field at the origin is therefore approximately 6107.83 N/C.

Answer:

Explanation:

Acceleration without friction on an inclined plane = g sinθ

Acceleration with friction on inclined plane = g sinθ - μ g cosθ

s = 1/2 a t²

For motion on friction-less surface

s = 1/2 g sinθ t₁²

For motion on frictional surface

s = 1/2( g sinθ - μ g cosθ)  t₂²

t₂ = 2t₁

(  sinθ - μ  cosθ)  t₂² =  sinθ t₁²

(  sinθ - μ  cosθ)  4t₁² =  sinθ t₁²

(  sinθ - μ  cosθ)  4 =  sinθ

4sinθ - sinθ = 4μ  cosθ

3sinθ = 4μ  cosθ

3 / 4 tanθ =μ  

μ  = .75  tanθ

Note: There is no image to take as a reference, so I'm assuming F2 directed to the right and F1 to the left, and F2=2F1

Answer:

[tex]\displaystyle a=\frac{F}{3M}[/tex]

to the right

Explanation:

Net Force

When several forces are applied to a particle or a system of particles, the net force is the sum of them all, considering each force as a vector. As for the second Newton's law, the total force equals the product of the mass by the acceleration of the system:

[tex]\vec F_n=m\cdot \vec a[/tex]

If the net force is zero, then the system of particles keeps at rest or at a constant velocity.

The system of particles described in the question consists of two objects of masses m1=M and m2, where

[tex]m_2=2m_1=2M[/tex]

Two forces F1=F and F2 act individually on each object in opposite directions and

[tex]F_2=2F_1=2F[/tex]

We don't get to see any image to know where the forces are applied to, so we'll assume F2 to the right and F1 to the left.

The net force of the system of particles is

[tex]F_n=2F-F=F[/tex]

The mass of the system is

[tex]m_t=m_1+m_2=3M[/tex]

Thus, the acceleration of the center of mass of the system is

[tex]\displaystyle a=\frac{F}{3M}[/tex]

Since F2 is greater than F1, the direction of the acceleration is to the right.

Note: If the forces were opposite than assumed, the acceleration would be to the left

The minimum coefficient of static friction between the pavement and the tires is 0.156.

The minimum coefficient of static friction is calculated by applying Newton's second law of motion in determining the net force.

[tex]F_c = Wsin(\theta) + \mu_s W cos(\theta)\\\\\frac{mv^2}{r} = mg sin(\theta) + \mu_s mg cos(\theta)\\\\\mu_s mg cos(\theta)\ = \frac{mv^2}{r} - mg sin(\theta) \\\\\mu _ s = \frac{mv^2 \ - \ mgr sin(\theta)}{mg rcos(\theta)}[/tex]

where;

[tex]\mu _ s = \frac{1200 \times 18.1^2 \ - \ 1200 \times 9.8 \times 100 sin(10.4)}{1200 \times 9.8 \times 100 \times cos(10.4)}\\\\\mu_s = 0.156[/tex]

Thus, the minimum coefficient of static friction between the pavement and the tires is 0.156.

Learn more about coefficient of static friction here: https://brainly.com/question/25050131

To calculate the resultant magnetic field at different points between the Helmholtz coils, use the formula B = (μ0 * N * I * R^2) / ((R^2 + x^2)^(3/2)). Plug in the given values to find the magnetic field at specific distances from the center of the coil.

To calculate the resultant magnetic field at different points, we can use the formula:

B = (μ0 * N * I * R^2) / ((R^2 + x^2)^(3/2))

Where B is the magnetic field, μ0 is the permeability of free space, N is the number of turns, I is the current, R is the radius of the coil, and x is the distance from the center of the coil.

Using this formula, we can calculate the magnetic field at x1 = 2.8 cm, x2 = 5.5 cm, x3 = 7.3 cm, and x4 = 11.0 cm by plugging in the given values.

Helmholtz coils create a uniform magnetic field using two identical coils with the same current. The formula for calculating the magnetic field at any position along the axis demonstrates this uniformity and the provided distances can be used to find specific magnetic field values.

A Helmholtz coil configuration consists of two identical circular coils, each having N turns, radius R, and carrying the same current I. The coils are separated by a distance equal to R. This setup creates a uniform magnetic field between the coils. Given data are R = 11.0 cm, I = 17.0 A, and N = 300 turns.

Using the formula for the magnetic field along the axis of Helmholtz coils:

This approach allows for the precise calculation of the magnetic field Bx at the given distances from the center, demonstrating the setup’s uniformity and reinforcement of the fields.

Complete Question : Two coils that are separated by a distance equal to their radius and that carry equal currents such that their axial fields add are called Helmholtz coils. A feature of Helmholtz coils is that the resultant magnetic field between the coils is very uniform. Let R = 11.0 cm, I = 17.0 A, and N = 300 turns for each coil. Place one coil in the y-z plane with its center at the origin and the other in a parallel plane at R = 11.0 cm. Calculate the resultant field Bx at x1 = 2.8 cm, x2 = 5.5 cm, x3 = 7.3 cm, and x4= 11.0 cm.

To solve this problem we will apply the concepts related to energy conservation. We know that potential energy is transformed into kinetic and vice versa energy. Since the energy accumulated in the upper part is conserved as potential energy, when the object is thrown all that energy will be converted into kinetic energy. Therefore we will have the following relation,

[tex]KE = PE[/tex]

[tex]KE = mgh[/tex]

Here,

m = mass

g = Gravitational acceleration

h = Height

Replacing,

[tex]KE = (6)(9.8)(5)[/tex]

[tex]KE = 294J[/tex]

Therefore the kinetic energy at the bottom is 294J